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This is motivated by idle curiosity. I recently learned a result of Duistermaat and Van Der Kallen in "Constant terms of powers of a Laurent polynomial" which says that:

If the constant term of $f^n$ vanishes for all $n\in \mathbb N$, where $f\in \mathbb C[x,x^{-1}]$ is a Laurent polynomial, then either $f\in \mathbb C [x]$ or $f\in \mathbb C[x^{-1}]$.

Their proof uses complex analysis and I was wondering if one can prove the statement over an arbitrary field of characteristic zero using algebraic or combinatorial methods. Characteristic zero is of course necessary since $x^{p-1}+\frac{1}{x}$ satisfies the conditions in characteristic $p$.

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    $\begingroup$ Our main result was about Laurent polynomials in several variables. $\endgroup$
    – Wilberd van der Kallen
    Commented Nov 15, 2011 at 16:01
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    $\begingroup$ Isn't it false without some convergence condition? Just take f=x^-2+g, where g= a1x+a2x^2+..., choose the a_m with m odd arbitrarily, and define the a_m with m even inductively to make the coefficients that you want equal to 0? $\endgroup$
    – paul Monsky
    Commented Nov 15, 2011 at 16:28
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    $\begingroup$ A polynomial has only finitely many terms, so you cannot define the a_m inductively like that. $\endgroup$
    – Wilberd van der Kallen
    Commented Nov 15, 2011 at 16:53
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    $\begingroup$ If $K$ is a field of characteristic $0$ and $f\in K[x,x^{-1}]$, then $f$ is also in $L[x,x^{-1}]$, where $L$ is the subfield of $K$ generated by the coefficients of $f$. As $L$ is of finite type over $\mathbb{Q}$, it can be embedded in $\mathbb{C}$, so you can see $f$ as an element of $\mathbb{C}[x,x^{-1}]$ and apply the result that you are citing. So the result that you want seems to at least be true. But I'm guessing that you wanted another proof. $\endgroup$
    – Alex
    Commented Nov 15, 2011 at 19:27
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    $\begingroup$ @Gjergji--There's one thing to be said for the algebraic proof. When we adjoin the solutions of f(x)=t to the field of fractions of K[[t]], we get the splitting field of a polynomial of degree r+s. When char K is >r+s this extension is tamely ramified. So each solution, a, still admits a Newton-Puiseux expansion, and the proof I gave in my answer still works---when char K is > r+s, then c1, c2,... cannot all be 0. As you pointed out there are counterexamples when char K = r+s. $\endgroup$
    – paul Monsky
    Commented Dec 28, 2011 at 20:03

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I'll give an algebraic argument, which is I think essentially the same as Duistermatt's, but substitutes partial fractions and some valuation theory for complex analysis. K will be algebraically closed of characteristic 0; f will be in K[x,1/x]. We suppose f is not in K[x] or K[1/x]. Let r and -s be the largest and smallest exponent of x appearing in f; r and s are >0.

Lemma:---Let M be a finite extension of the field of fractions of K[[t]]. Extend the obvious valuation on the field of fractions to M. Then if a is in M with f(a)=t, ord f'(a) must be < 1. (To see this note that ord a is 0. Then a has a Newton-Puiseux expansion a0 +(a1)(t^p)+..., with a0 and a1 non-zero, and p a positive rational. The derivation D=d/dt extends to M, and 1=D(f(a)) is the product of f'(a) by (a1)(p)*(t^(p-1))+... So ord f'(a)=1-p which is < 1.)

Theorem 1:---Let S be a subset of the algebraic closure of K(z). Suppose that for each a in S, f(a)=1/z. Then the sum over S of the 1/(af'(a)) cannot be z. (To see this let t=1/z. Then K(z)= K(t) which imbeds in the field of fractions of K[[t]], and we may view S as a subset of a finite extension, M, of this field. By the lemma, each 1/(af'(a)) has ord > -1. But z=1/t has ord equal to -1.)

Now let c_n be the constant term in f^n, and W=1+... be the element sigma (c_n)(z^n) of K[[z]]. Combinatorialists know that a partial fraction argument shows that W is algebraic over L=K(z). Carrying out the partial fraction argument explicitly one finds:

Theorem 2:---There are a lying in a finite extension of L with each f(a) equal to 1/z, such that zW is the sum of the 1/(af'a)). (So by Theorem 1, W is not 1, and c_1, c_2,... cannot all be 0.)

I'll sketch a proof of Theorem 2. Let U be the element (x^s)(1-zf) of L[x]. If a is a root of u, then f(a)=1/z. So by the lemma f'(a) is not 0. Then U'(a) is not 0, and U is separable. It follows that 1/(1-fz)=(x^s)/U is a sum of c/(x-a) where a runs over the roots of f(x)=1/z. It's easy to see that the c corresponding to a given a is -1/(zf'(a)). I now refer to Lemmas 3.6 and 3.7 of my article "Generating functions attached to some infinite matrices"---see arXiv 0906.1836 or Elec. J. of Comb, v.18 (1) 2011:

If we take U1=x^s and U2=U=(x^s)(1-zf) we are in the situation of Lemma 3.7. W is "the coefficient of x^0 in the element (U1)/(U2)". In the language of the lemma, W is the sum of the l_0(c/(z-a)). The proof of Lemma 3.6 shows that l_0(1/(z-a)) is either 0 or -1/a. So z l_0(c/(z-a)) is either 0 or 1/(af'(a)). This completes the proof.

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